In the previous section we saw theorem 27.9 and it's converse. In this section we will see some solved examples.
Solved example 27.15
In the fig.27.46(a) below, A, B, C and D are four points on a circle.
∠DBC = 55o and ∠CAB = 45o. Compute ∠BCD
Solution:
In the fig.27.46(a) below, A, B, C and D are four points on a circle.
AC and BD intersect at E in such a way that ∠BEC = 30o. and ∠ECD = 20o. Find ∠BAC
Solution:
1. ∠BEC and ∠BEA form a linear pair. So ∠BEA = 180 - ∠BEC = 180 -130 = 50o.
2. ∠BEA and ∠CED are opposite angles, and are hence equal. So we get ∠BEA = ∠CED = 50o.
Solved example 27.15
In the fig.27.46(a) below, A, B, C and D are four points on a circle.
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| Fig.27.45 |
Solution:
1. Consider arc BC in fig(b). It subtends ∠BAC (= 45o) on the alternate arc.
• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 45o. This is marked in fig(b)
2. Consider ΔBDC. We get ∠BCD = [180-(45+55)] =[180-100] = 80o. (∵ sum of interior angles of a triangle is 180o)
Thus we get the required angle. We can do a check by using theorem 27.9.
Solved examples 27.162. Consider ΔBDC. We get ∠BCD = [180-(45+55)] =[180-100] = 80o. (∵ sum of interior angles of a triangle is 180o)
Thus we get the required angle. We can do a check by using theorem 27.9.
3. Consider arc CD in fig(c). It subtends ∠CBD (= 55o) on the alternate arc.
• The same arc subtends ∠CAD on the alternate arc. So ∠CBD = ∠CAD = 55o. This is marked in fig(c)
(i) Sum of opposite angles ∠BAD and ∠BCD = 45 + 55 + 80 = 180o.
(i) Sum of opposite angles ∠BAD and ∠BCD = 45 + 55 + 80 = 180o.
In the fig.27.46(a) below, A, B, C and D are four points on a circle.
 |
| Fig.27.46 |
Solution:
1. ∠BEC and ∠BEA form a linear pair. So ∠BEA = 180 - ∠BEC = 180 -130 = 50o.
2. ∠BEA and ∠CED are opposite angles, and are hence equal. So we get ∠BEA = ∠CED = 50o.
3. Consider ΔCED. We get ∠EDC = [180-(50+20)] =[180-70] = 110o. (∵ sum of interior angles of a triangle is 180o)
Solved example 27.17
In the fig.27.47(a) below, ABCD is a square.
Determine ∠DPC
Solution:
1. Draw the diagonal AC of the square.
2. A diagonal of a square will bisect the angles at the corners. So we get:
∠DAC = ∠BAC = 45o.
3. Consider the quadrilateral ACPD. It is a cyclic quadrilateral. The sum of opposite angles = 180o.
4. So we get: ∠DPC + ∠DAC = 180o ⇒ ∠DPC + 45 = 180 ⇒ ∠DPC = 180 - 45 = 135o.
Now we will see an important result related to cyclic quadrilaterals. We will learn it in steps:
4. Consider the major arc BC in fig(c). It subtends ∠BDC (= 110o) on the alternate arc.
• The same arc subtends ∠BAC on the alternate arc. So ∠BAC = ∠BDC = 110o. This is marked in fig(c). Thus we get the required angle.Solved example 27.17
In the fig.27.47(a) below, ABCD is a square.
 |
| Fig.27.47 |
Solution:
1. Draw the diagonal AC of the square.
2. A diagonal of a square will bisect the angles at the corners. So we get:
∠DAC = ∠BAC = 45o.
3. Consider the quadrilateral ACPD. It is a cyclic quadrilateral. The sum of opposite angles = 180o.
4. So we get: ∠DPC + ∠DAC = 180o ⇒ ∠DPC + 45 = 180 ⇒ ∠DPC = 180 - 45 = 135o.
1. Fig.27.48(a) below shows a cyclic quadrilateral ABCD.
2. The side AB is extended along towards the right up to point E. So ∠CBE (shown in red colour) becomes an exterior angle of the cyclic quadrilateral. We can write:
■ ∠CBE is the exterior angle of the cyclic quadrilateral ABCD at the vertex B.
3. For the vertex B, the opposite vertex is D
• So, for the vertex B, ∠ADC (shown in yellow colour) is the 'interior angle at the opposite vertex'
4. Thus we have three quantities:
(i) A vertex B (ii) Exterior angle at that vertex B (iii) Interior angle at D, which is the opposite vertex of B
• We want to know the relation between (ii) and (iii)
5. Consider the interior angle at B. It is shown in blue colour in fig (b)
• Blue + Red will obviously be 180o (∵ they form a linear pair)
• So we can write: ∠CBE + ∠ABC = 180o
6. Yellow and Blue are opposite angles of a cyclic quadrilateral. So their sum will be 180o.
• We can write: ∠ADC + ∠ABC = 180o
7. From (5) we get: ∠ABC = 180 – ∠CBE
• Substituting this in (6) we get:
∠ADC + (180 – ∠CBE) = 180
⇒ ∠ADC – ∠CBE = 180 – 180
⇒ ∠ADC – ∠CBE = 0
⇒ ∠ADC = ∠CBE
8. So we can write:
• The exterior angle at B is equal to the interior angle at opposite vertex
9. Now consider fig (c). The side CB is extended upto F
• ∠ABF is an exterior angle at vertex B. So is ∠CBE
• But we can see that the two are equal because, they are opposite angles.
• So, at a vertex, there will be only one value for an exterior angle
10. We can write the above results in a general form:
• Consider any vertex of a cyclic quadrilateral.
• There will be an exterior angle at that vertex
• That exterior angle will be equal to the interior angle at the opposite vertex
Some solved examples on cyclic quadrilaterals are shown in the form of a video presentation at the following links:
Trapezium Cyclic or not
Non-rectangular parallelogram Cyclic or not
 |
| Fig.27.48 |
■ ∠CBE is the exterior angle of the cyclic quadrilateral ABCD at the vertex B.
3. For the vertex B, the opposite vertex is D
• So, for the vertex B, ∠ADC (shown in yellow colour) is the 'interior angle at the opposite vertex'
4. Thus we have three quantities:
(i) A vertex B (ii) Exterior angle at that vertex B (iii) Interior angle at D, which is the opposite vertex of B
• We want to know the relation between (ii) and (iii)
5. Consider the interior angle at B. It is shown in blue colour in fig (b)
• Blue + Red will obviously be 180o (∵ they form a linear pair)
• So we can write: ∠CBE + ∠ABC = 180o
6. Yellow and Blue are opposite angles of a cyclic quadrilateral. So their sum will be 180o.
• We can write: ∠ADC + ∠ABC = 180o
7. From (5) we get: ∠ABC = 180 – ∠CBE
• Substituting this in (6) we get:
∠ADC + (180 – ∠CBE) = 180
⇒ ∠ADC – ∠CBE = 180 – 180
⇒ ∠ADC – ∠CBE = 0
⇒ ∠ADC = ∠CBE
8. So we can write:
• The exterior angle at B is equal to the interior angle at opposite vertex
9. Now consider fig (c). The side CB is extended upto F
• ∠ABF is an exterior angle at vertex B. So is ∠CBE
• But we can see that the two are equal because, they are opposite angles.
• So, at a vertex, there will be only one value for an exterior angle
10. We can write the above results in a general form:
• Consider any vertex of a cyclic quadrilateral.
• There will be an exterior angle at that vertex
• That exterior angle will be equal to the interior angle at the opposite vertex
Trapezium Cyclic or not
Non-rectangular parallelogram Cyclic or not
In the next section, we will learn about Multiplication of Chords.
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